Dicuss the continuity of the function:
`f(x)={{:(1+x^(2)", "0lexle1),(2-x", "xgt1):}` at x = 1.

Answer

To discuss the continuity of the function

\[ f(x) = \begin{cases} 1 + x^2 & \text{for } 0 \leq x \leq 1 \\ 2 - x & \text{for } x > 1 \end{cases} \]

at \( x = 1 \), we need to check the following conditions:

1. **Value of the function at \( x = 1 \)**:
\[ f(1) = 1 + 1^2 = 1 + 1 = 2 \]

2. **Left-hand limit as \( x \) approaches 1**:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (1 + x^2) = 1 + 1^2 = 2 \]

3. **Right-hand limit as \( x \) approaches 1**:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2 - x) = 2 - 1 = 1 \]

4. **Check continuity**:
For \( f(x) \) to be continuous at \( x = 1 \), the following must hold:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \]
From our calculations:
- \( f(1) = 2 \)
- \( \lim_{x \to 1^-} f(x) = 2 \)
- \( \lim_{x \to 1^+} f(x) = 1 \)

Since the left-hand limit (2) does not equal the right-hand limit (1), we conclude that:

\[ \text{The function } f(x) \text{ is discontinuous at } x = 1. \]